## Introduction

Projective Geometry is a field of mathematics most commonly used for anything to do with cameras - rendering, 3d reconstruction etc. I've only a weak knowledge of it, but this method of calculating the intersection of two lines falls out of the first chapter or two in a standard textbook. I found it very surprising, and hopefully you find it interesting, too!

First off - a demo of the code itself:

import numpy as np

def intersectLines(s1, e1, s2, e2):  # start of line 1/ end of line 1 etc
# Append 1 to all points. Make them (x, y, 1)
s1, e1, s2, e2 = [np.append(p, 1) for p in [s1, e1, s2, e2]]
# Solve - just cross product everything (!)
kx, ky, k = np.cross(np.cross(s1, e1), np.cross(s2, e2))
return ([kx/k, ky/k] if k != 0 else None)

If you're thinking 'why does that work???', or 'where's the bit with the equations?' you're not alone! Cross products are usually only seen when trying to find orthogonal vectors. Here we're appending a 1 to some points, cross producting them twice, and somehow ending up with the intersection of the lines.

Let's start off by working out the inner cross products: np.cross(s1, e1) and np.cross(s2, e2). Each are simply taking the product of two points which both lie along the same line.

### Points -> Lines

Here we'll show that these products calculate a line equation. Consider two points $\vec{S}$ and $\vec{E}$.

$\begin{bmatrix} S_x \\ S_y \\ 1 \end{bmatrix} \ \times \ \begin{bmatrix} E_x \\ E_y \\ 1 \end{bmatrix} \ = \ \begin{bmatrix} S_y - E_y \\ E_x - S_x \\ S_xE_y - S_yE_x \end{bmatrix} \ = \ \begin{bmatrix} a \\ b \\ c \end{bmatrix} \$

This first cross product actually gives the equation of the line defined by $\vec{S}$ and $\vec{E}$ in the form $ax + by + c = 0$. Let's quickly verify…

\begin{align} ax + by + c = 0 \\ (S_y - E_y)x + (E_x - S_x)y + S_xE_y - S_yE_x = 0 \end{align}

Verifying $\vec{S}$ lies on the line $ax + by + c = 0$:

\begin{align} \require{cancel} (S_y - E_y)S_x + (E_x - S_x)S_y + S_xE_y - S_yE_x = 0 \\ S_xS_y - S_xE_y + S_yE_x - S_xS_y + S_xE_y - S_yE_x = 0 \\ \cancel{S_xS_y - S_xS_y} + \cancel{S_xE_y - S_xE_y} + \cancel{S_yE_x - S_yE_x} = 0 \\ 0 = 0 \end{align}

Verifying $\vec{E}$ lies on the line $ax + by + c = 0$:

\begin{align} (S_y - E_y)E_x + (E_x - S_x)E_y + S_xE_y - S_yE_x = 0 \\ S_yE_x - E_xE_y + E_xE_y - S_xE_y + S_xE_y - S_yE_x = 0 \\ \cancel{S_yE_x - S_yE_x} + \cancel{E_xE_y - E_xE_y} + \cancel{S_xE_y - S_xE_y} = 0 \\ 0 = 0 \end{align}

Hence $ax + by + c = 0$ is an equation of the line passing through $\vec{S}$ and $\vec{E}$.

### Lines -> Intersection

The cross product of two line equations gives their intersection… Let's check.

Take two lines: $a_1x + b_1y + c_1 = 0$ defined by $\begin{bmatrix} a_1 \\ b_1 \\ c_1 \end{bmatrix}$ and $a_2x + b_2y + c_2 = 0$ defined by $\begin{bmatrix} a_2 \\ b_2 \\ c_2 \end{bmatrix}$.

$\begin{bmatrix} a_1 \\ b_1 \\ c_1 \end{bmatrix} \times \begin{bmatrix} a_2 \\ b_2 \\ c_2 \end{bmatrix} \ = \ \begin{bmatrix} b_1c_2 - c_1b_2 \\ c_1a_2 - a_1c_2 \\ a_1b_2 - b_1a_2 \end{bmatrix} \$ Now, convert to the $\begin{bmatrix}x \\ y \\ 1 \end{bmatrix}$ format. This means we must divide by $a_1b_2 - b_1a_2$. We quickly check this is non-zero. Assuming the lines have an intersection $\implies$ their slopes are different $\implies$ $\frac{a_1}{b_1} \neq \frac{a_2}{b_2} \implies \frac{a_1}{b_1} - \frac{a_2}{b_2} \neq 0 \implies a_1b_2 - b_1a_2 \neq 0$. Conversely if $a_1b_2 - b_1a_2 = 0$ then the lines are parallel and have no intersection.

$\begin{bmatrix} P \\ 1 \end{bmatrix} = \begin{bmatrix} (b_1c_2 - c_1b_2)/(a_1b_2 - b_1a_2) \\ (c_1a_2 - a_1c_2)/(a_1b_2 - b_1a_2) \\ 1 \end{bmatrix} \$ Finally the vector $P \in \mathbb{R}^2$ is the first values of this vector, and is the intersection of the lines! It's worth noting in projective geometry the point $(x, y, 1)$ is equilivant to $(kx, ky, k)$

To quickly verify this:

$P$ lies on $a_1x + b_1y + c_1 = 0$:

\begin{align} a_1\frac{b_1c_2 - c_1b_2}{a_1b_2 - b_1a_2} + b_1\frac{c_1a_2 - a_1c_2}{a_1b_2 - b_1a_2} + c_1 = 0 \\ a_1(b_1c_2 - c_1b_2) + b_1(c_2a_2 - a_1c_2) + c_1(a_1b_2 - b_1a_2) = 0 \\ a_1b_1c_2 - a_1c_1b_2 + b_1c_1a_2 - a_1b_1c_2 + a_1c_1b_2 - b_1c_1a_2 = 0 \\ \cancel{a_1b_1c_2 - a_1b_1c_2} + \cancel{a_1c_1b_2 - a_1c_1b_2} + \cancel{b_1c_1a_2 - b_1c_1a_2} = 0 \\ 0 = 0 \end{align}

In the same way it's straightforward to show $P$ lies on $a_2x + b_2y + c_2 = 0$. Hence $P$ is an intersection of the lines (provided one exists)!

### Sample code in C#

This is a handy snippet to have for Unity. It's worth noting, however, that this isn't the most numerically stable or fastest way I'm sure. I use it because it's short, neat and I don't require any higher accuracy or speed from it.

private static bool TryIntersectLines(Vector2 start1, Vector2 end1, Vector2 start2, Vector2 end2, out Vector2 intersection) {
// Append 1 to all points. (x, y) -> (x, y, 1)
Vector3 toProjPlane(Vector2 p) => new Vector3(p.x, p.y, 1);

// Get equations of lines using cross product & solve
Vector3 l1 = Vector3.Cross(toProjPlane(start1), toProjPlane(end1));
Vector3 l2 = Vector3.Cross(toProjPlane(start2), toProjPlane(end2));
Vector3 sol = Vector3.Cross(l1, l2);

if (sol.z == 0) {
// No solution - result 'at infinity' in projective space.
intersection = default(Vector2);
return false;
}
else {
intersection = new Vector2(sol.x/sol.z, sol.y/sol.z);
return true;  //** See below!
}
}

### A note on line segments

The previous examples only deal with the intersection of lines defined by two points on them. If you require intersection of two line segments defined by start and end points, you'll need an extra condition. Checking if the intersection is bounded in x and y by the ends of each line is sufficient - e.g.

// return true;  // <-- replace this with...
// Check the intersection is within the boundaries of both lines
return (
(intersection.x < start1.x) != (intersection.x < end1.x)     // Bounded in x by line 1
&& (intersection.y < start1.y) != (intersection.y < end1.y)  // Bounded in y by line 1
&& (intersection.x < start2.x) != (intersection.x < end2.x)  //  ...            line 2
&& (intersection.y < start2.y) != (intersection.y < end2.y)  //  ...            line 2
);